Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 2e5 + 4;
int l[N], r[N];
int main() {
  int sy, sa, sb;
  scanf("%d", &sy);
  scanf("%d", &sa);
  scanf("%d", &sb);
  int n;
  scanf("%d", &n);
  int len[n];
  for (int i = 0; i < n; i++)
    scanf("%d", &l[i]), scanf("%d", &r[i]), len[i] = r[i] - l[i];
  for (int i = 1; i < n; i++) len[i] += len[i - 1];
  int q;
  scanf("%d", &q);
  while (q--) {
    int px, py;
    scanf("%d", &px);
    scanf("%d", &py);
    long long den = py - sy;
    assert(den > 0);
    long long xl = (-1LL * sy * px + 1LL * sa * py);
    xl = (xl + den - 1) / den;
    long long xr = (1LL * sb * py - 1LL * px * sy);
    xr /= den;
    int al, ar;
    ar = upper_bound(r, r + n, xr) - r;
    ar--;
    al = lower_bound(l, l + n, xl) - l;
    long double ans = 0.0;
    if (ar >= al) {
      ans = len[ar];
      if (al - 1 >= 0) ans -= len[al - 1];
    }
    al--;
    ar++;
    long double ldxl, ldxr;
    ldxl = (long double)(-1LL * sy * px + 1LL * sa * py) / (long double)den;
    ldxr = (long double)(1LL * sb * py - 1LL * px * sy) / (long double)den;
    int flag = 0;
    if (al >= 0 and ar < n) {
      long double hr = r[al], hl = l[ar];
      if (hr >= ldxr or hl <= ldxl) ans += (ldxr - ldxl), flag = 1;
    }
    if (flag == 0) {
      if (al >= 0) {
        if ((long double)r[al] >= ldxl) ans += (r[al] - ldxl);
      }
      if (ar < n) {
        if ((long double)l[ar] <= ldxr) ans += (ldxr - l[ar]);
      }
    }
    ans *= (long double)(py - sy);
    ans /= (long double)py;
    cout << fixed << setprecision(10) << ans << "\n";
  }
  return 0;
}